Slopes of perpendicular lines — lesson. Mathematics State Board, Class 10.

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Let

$l_1$ and

$l_2$ be two non-vertical lines.

The slope of line

$l_1$ is

$m_1$ , and line

$l_2$ is

$m_2$ .

Let the inclination of

$l_1$ be

$\theta_1$ and

$l_2$ be

$\theta_2$ .

Then,

$m_1 = \tan \theta_1$ and

$m_2 = \tan \theta_2$ .

Assume

$l_1$ and

$l_2$ are perpendicular lines.

$\Delta ABC$ :

$\angle A = \theta_1$ and

$\angle C = 90^\circ$

Then,

$\angle B = 180^\circ - \angle A - \angle C$ [by Angle sum property].

$\angle B = 180^\circ - \theta_1 - 90^\circ$

$\angle B =90^\circ - \theta_1$

Measuring slope of

$l_2$ through angle

$\theta_2$ and

$90^\circ - \theta_1$ are opposite of each other.

$\Rightarrow \tan \theta_2 = - \tan (90^\circ - \theta_1)$

$\Rightarrow \tan \theta_2 = - \frac{\sin(90^\circ - \theta_1)}{\cos(90^\circ - \theta_1)}$

[Using the trigonometric identities

$\sin (90^\circ - A) = \cos A$ and

$\cos (90^\circ - A) = \sin A$ ]

$\Rightarrow \tan \theta_2 = - \frac{\cos \theta_1}{\sin \theta_1}$

$\Rightarrow \tan \theta_2 = - \cot \theta_1$

$\Rightarrow \tan \theta_2 = - \frac{1}{\tan \theta_1}$

$\Rightarrow \tan \theta_1 \cdot \tan \theta_2 = -1$

$\Rightarrow m_1 \cdot m_2 = -1$

Conversely:

Let

$l_1$ and

$l_2$ be two non-vertical lines with slopes

$m_1$ and

$m_2$ , respectively, such that

$m_1 \cdot m_2 = -1$ .

$\Rightarrow \tan \theta_1 \cdot \tan \theta_2 = -1$

$\Rightarrow \tan \theta_1 = - \frac{1}{\tan \theta_2}$

$\Rightarrow \tan \theta_1 = - \cot \theta_2$

$\Rightarrow \tan \theta_1 = - \tan (90^\circ - \theta_2)$ [Using

$\cot (90^\circ - A) = \tan A$ ]

$\Rightarrow \tan \theta_1 = \tan (-(90^\circ - \theta_2))$ [Using

$\tan (-A) = - \tan A$ ]

$\Rightarrow \tan \theta_1 = \tan (\theta_2 - 90^\circ)$

$\Rightarrow \theta_1 = \theta_2 - 90^\circ$ [since

$0^\circ \le \theta_1, \theta_2 \le 180^\circ$ ]

$\Rightarrow \theta_2 = \theta_1 + 90^\circ$ - - - - - (I)

An exterior angle of a triangle is equal to the sum of two opposite interior angles.

$\Rightarrow \theta_2 = \angle A + \angle C$

$\Rightarrow \theta_2 = \theta_1 + \angle C$ - - - - - (II)

On comparing (I) and (II), we get:

$\angle C = 90^\circ$

Thus, the lines

$l_1$ and

$l_2$ are perpendicular.

Two lines are perpendicular if and only if the slopes are negative reciprocal of each other. That is,

$m_1 \cdot m_2 = -1$ .

Example:

The slope of the line

$p$ is

$\frac{- 20}{5}$ , and the slope of the line

$q$ is

$\frac{8}{32}$ . Are the lines

$p$ and

$q$ are perpendicular?

Solution:

Let

$m_1$ be the slope of

$p$ , and

$m_2$ be the slope of

$q$ .

$m_1 \cdot m_2 = -1$ , then the lines are perpendicular.

$m_1 =$

$\frac{- 20}{5}$

$=$

$-4$

$m_2 =$

$\frac{8}{32}$

$=$

$\frac{1}{4}$

$m_1 \cdot m_2 = - 4 \ \times$

$\frac{1}{4}$

$m_1 \cdot m_2 = - 1$

Therefore,

$p$ and

$q$ are perpendicular lines.

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5. Slopes of perpendicular lines

Theory: